# Dual vector space properties ($\forall \phi\in E^*,\langle\phi,x\rangle=0\implies x=0$)

$E$ is a finite-dimensional vector space over a field K and $E^∗$ its dual.

$$\boxed{x\in E, x=0\iff \forall \phi\in E^*,\langle\phi,x\rangle=0}$$

The proof for ($\implies$) seems to be obvious since $x=0\implies\langle\phi,x\rangle=\phi(x)=0$

But for ($\Longleftarrow$)that is a little more difficult, below the proof:

$\bigg [\forall \phi\in E^∗,\langle \phi,x\rangle=0\implies x=0\bigg ]\iff \bigg[x\in E, x\neq0\implies \exists \phi \in E^*, \; \langle\phi,x\rangle\neq 0\bigg]$

Let $x\neq 0$ and $\mathcal{B}=\big\{e_1,e_2,...e_n\big\} \text{a basis of$E$} \implies x=\lambda_1e_1+\lambda_2e_2+...+\lambda_ie_i+...+\lambda_ne_n\quad i\in[\![0,n]\!]$ so there exists a basis of $E$ such that $\mathcal{B}'=\big\{x,u_2,u_3,..u_n\big\}$

Let $f$ such that:

$\begin{array}lf:&E&\to K\\&\lambda_iu_i&\to \lambda_i\end{array}\qquad\implies f\in E^* \text{ and } f(x)=1$

If the proof is correct, I understand the logical path, but I don't really understand it?

I know there is a similar question but it doesn't help me: Can a non-zero vector have zero image under every linear functional?

• What does $\langle \phi, x \rangle$ mean? $\phi$ is in the dual and is a function Commented Aug 8, 2017 at 22:19
• @tattwamasiamrutam Probably the action of $\phi$ on $x$.
– user296602
Commented Aug 8, 2017 at 22:20
• $\begin{array}lu:&E^*&\to K\\&\phi&\to \langle\phi,x\rangle\end{array}$
– Stu
Commented Aug 8, 2017 at 22:27

Let $dim(E)=n$ and $A=\{v_1,v_2...v_n\}$ a basis of $E$.

Now let $x \in E$,then $x=a_1v_n+a_2v_2+...a_nv_n$ for some $a_1...a_n \in F$

Take the set of linear functionals $B=\{f_1,f_2...f_n\}$ where $f_i(v_j)=0$ if $i \neq j$ and $f_i(v_j)=1$ if $i=j$.

We have that $$0=f_i(x)=a_1f_i(v_1)+a_2f_i(v_2)+...+a_if_i(v_i)+...+a_nf_i(v_n)=a_i,\forall i \in \{1,2....n\},$$

Thus $a_i=0,\forall i \in \{1,2....n\} \Rightarrow x= \bar{0}$

• So, If I understood, you are proving by contrapostion, $x\in E, x\neq0\implies \exists \phi \in E^*, \; \langle\phi,x\rangle\neq 0$
– Stu
Commented Aug 8, 2017 at 22:41
• @Stu..No i just took an appropriate collection of functionals because of the hypothesis in your statement to prove that x=0..If you want you can prove that these maps are indeed linear functionals.An $f_i$ acts on $x$ and gives you the $i$-th coefficient of the linear combination of $x$ Commented Aug 8, 2017 at 22:44
• But also if you want you can try proving your statement by contraposition using the same argument Commented Aug 8, 2017 at 22:49
• I will try by contraposition, see you soon!! and thanks
– Stu
Commented Aug 8, 2017 at 22:50
• you are welcome Commented Aug 8, 2017 at 22:51

$\bigg [\forall \phi\in E^∗,\langle \phi,x\rangle=0\implies x=0\bigg ]\iff \bigg[x\in E, x\neq0\implies \exists \phi \in E^*, \; \langle\phi,x\rangle\neq 0\bigg]$

Let $x\neq 0$ and $\mathcal{B}=\big\{e_1,e_2,...e_n\big\} \text{a basis of$E$} \implies x=\lambda_1e_1+\lambda_2e_2+...+\lambda_ie_i+...+\lambda_ne_n\quad i\in[\![0,n]\!]$ so there exists a basis of $E$ such that $\mathcal{B}'=\big\{x,u_2,u_3,..u_n\big\}$

Let $f$ such that:

$\begin{array}lf:&E&\to K\\&\lambda_iu_i&\to \lambda_i\end{array}\implies f(x)=1$

Is it correct?

• can you explain a little bit more your ideas of your solution? Commented Aug 9, 2017 at 11:17
• By contraposition we have to find one $\phi$ such that $\forall x\neq0 \implies \phi(x)\neq 0$,since $x\neq 0 \implies \{x\}$ is linearly independant, so we can complete this set to obtain a basis of E. Then that remains to find a linear form $f$ such that $f(x)\neq 0$. But my problem is that I'm surprised by the result, indeed if I understood, this proof means for all linear forms $\phi(x)=0\implies x=0$
– Stu
Commented Aug 9, 2017 at 11:47
• i believe that contrapositive is that we assume that exists a $x \neq 0$ and from this we prove that exists a functional $f$ such that $f(x) \neq 0$ Commented Aug 9, 2017 at 12:03
• you can also prove this simply by contradiction if you want Commented Aug 9, 2017 at 12:07
• Ineed its quite a nice resulty the statement of this theorem..and this theorem can be applied also to infinite dimensional spaces which are equiped with a norm and the topology that the norm gives to such a space(and to finite dimensional spaces with a norm also).This is a result of the Hahn-Banach Theorem in functional analysis Commented Aug 9, 2017 at 12:19

Hint:

Choose a basis $(e_1,\dots, e_n)$ in $E$, and consider the dual basis $(e_1^*,\dots e_n^*)$ in $E^*$. Each $e_i^*$ is just the $i$-th coordinate map relative to the chosen basis in $E$.